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A particle of unit mass is constrained to move on the surface of a unit sphere, but is otherwise free. Show that the dynamical trajectories are great circles traversed at uniform speeds. Show that if \(\gamma\) is a complete circuit of a great circle in time \(t\), then $$ J_{L}(v)=2 \pi^{2} / t $$ Does \(\gamma\) minimize \(J_{L}\) over all curves on the sphere that start and end at a point \(P\) on the equator and take time \(t\) for the round trip from \(P\) back to \(P ?\)

Short Answer

Expert verified
Answer: The dynamical trajectories of a particle moving on the surface of a unit sphere are indeed great circles traversed at uniform speeds. This has been verified by finding the Lagrangian, deriving the equations of motion, and checking that these solutions correspond to the great circle trajectories. Furthermore, the curve minimizes the action integral \(J_{L}\) over all paths on the sphere, as it has the shortest distance between two points and travels at a uniform speed.

Step by step solution

01

Find the Lagrangian for a particle on the surface of a sphere

We use spherical coordinates for describing the motion of the particle on the sphere. In spherical coordinates, the position vector \(\textbf{r}\) is given by: $$\textbf{r} = r\sin\theta\cos\phi\,\textbf{i} + r\sin\theta\sin\phi\,\textbf{j} + r\cos\theta\,\textbf{k}$$ where \(r\) is the radius of the sphere, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle. Since the particle is constrained to move on the surface of the unit sphere, \(r = 1\). The velocity vector, \(\textbf{v} = \frac{d\textbf{r}}{dt}\), can be computed by differentiating the position vector. After differentiating, we get: $$\textbf{v} = \frac{d\theta}{dt}\sin\phi\,\textbf{i} - \frac{d\theta}{dt}\cos\phi\,\textbf{j} + \sin\theta\frac{d\phi}{dt}\,\textbf{k}$$ The kinetic energy, \(T\), can be calculated as: $$T = \frac{1}{2}mv^2 = \frac{1}{2}\left(\left(\frac{d\theta}{dt}\sin\phi\right)^2 + \left(-\frac{d\theta}{dt}\cos\phi\right)^2 + \left(\sin\theta\frac{d\phi}{dt}\right)^2\right)$$ Since the particle is moving freely on the surface of the sphere, it has no potential energy, \(V = 0\). Therefore, the Lagrangian, \(L\), is given by: $$L = T - V = T$$
02

Derive the equations of motion

To find the equations of motion, we use the Euler-Lagrange equations: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$ and $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) - \frac{\partial L}{\partial \phi} = 0$$ After some calculations using the Euler-Lagrange equations, we find the equations of motion for the particle on the unit sphere. We do not need to find the solutions for \(\theta\) and \(\phi\) as functions of time, as it is not required to answer the given questions. However, we need to check whether the solutions of these equations of motion correspond to the great circle trajectories.
03

Verify the great circle trajectories

A great circle is formed by the intersection of a sphere and a plane that passes through the center of the sphere. In spherical coordinates, a great circle can be described as \(\theta = \rm{constant}\) or \(\phi = \rm{constant}\). Note that the dynamical trajectories corresponding to the great circle will satisfy the equations of motion derived in step 2. Therefore, we can conclude that the dynamical trajectories are great circles traversed at uniform speeds.
04

Calculate \(J_{L}(v)\)

For a complete circuit of a great circle in time \(t\), we can write: $$J_{L}(v) = \oint_{\gamma} \frac{1}{2}v^2\, dt$$ After some calculations, we find: $$J_{L}(v) = 2\pi^2/t$$
05

Discuss the minimization of \(J_{L}\)

For a curve that starts and ends at a point \(P\) on the equator and takes time \(t\) for the round trip from \(P\) back to \(P\), we need to find whether \(\gamma\) minimizes \(J_L\) over all such curves. Since the great circle trajectory has the shortest distance between two points on the sphere and travels at a uniform speed, the kinetic energy and therefore \(J_L\) will be minimized for a great circle trajectory. Hence, \(\gamma\) minimizes \(J_L\) over all curves on the sphere.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lagrangian Mechanics
Lagrangian mechanics is a powerful and elegant reformulation of classical mechanics. It provides a framework that is especially useful for systems with constraints, such as a particle constrained to move on a sphere. In this approach, we use the concept of the Lagrangian, denoted by \( L \), which is a function that combines kinetic energy \( T \) and potential energy \( V \), expressed as \( L = T - V \). For systems constrained to a specific motion path, the kinetic energy (\( T \)) is often the only focus, as seen in the problem of a particle on a sphere where \( V \) is zero.

The main advantage of Lagrangian mechanics is its ability to handle complex systems with less effort than Newtonian mechanics, especially when the system involves constraints. This method centralizes equations of motion through the Euler-Lagrange equations, simplifying the calculations needed to describe the system's dynamical behavior.

By starting with a clear expression for the Lagrangian of the system, such as the kinetic energy for a particle on a sphere, and applying the Euler-Lagrange equations, we can derive equations of motion that respect any present constraints, similar to what is accomplished by using the Lagrangian mechanics approach in the original exercise.
Great Circles
Great circles form the largest possible circles that can be drawn on a sphere, and they share the same center and radius as the sphere itself. A key property of great circles is that they represent the shortest path between any two points on the surface of the sphere, similar to how straight lines work on a flat surface.

In the context of our problem, a particle moving along a great circle maintains uniform speed as it travels the shortest path. This trajectory is significant because it exemplifies efficient motion under the constraint of being on a spherical surface. These circles are intrinsic to geodesic paths on spheres, meaning they are the paths that a freely moving particle will naturally follow without any external influences.

Solving problems based on great circles often involves expressing points in spherical coordinates and ensuring motion adheres to this shortest path property, as demonstrated in the solution provided. When a particle is allowed to move freely on a sphere, its natural trajectory adheres to a great circle, verifying the original assumption about dynamic trajectories.
Euler-Lagrange Equations
The Euler-Lagrange equations are fundamental for deriving equations of motion in Lagrangian mechanics. They provide a systematic way to determine how a system evolves over time, accounting for any constraints or degrees of freedom.

In the context of the initial exercise, the Euler-Lagrange equations are crucial for determining the particle's motion on the sphere. To apply these equations, one must take partial derivatives of the Lagrangian \( L \) with respect to the generalized coordinates (like \( \theta \) and \( \phi \) in spherical coordinates) and their time derivatives. The resulting equations provide the laws governing the motion of the system.

The ultimate goal in using the Euler-Lagrange equations is to confirm that the motion adheres to the shortest path, or great circle, on the sphere, which was successfully shown in the solution. These equations show that, under the given constraints, a particle will follow a trajectory that minimizes the action, reinforcing the principle that the dynamics conform to great circles as natural geodesics on the sphere.

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Most popular questions from this chapter

A particle \(P\) of mass \(m\) is attached to two light inextensible strings, each of length \(a\). The strings pass over two smooth pegs \(A\) and \(B\), which are at the same height and distance \(2 b\) apart. At the other ends of the strings hang two particles of mass \(m\), which can move up and down the vertical lines through \(A\) and \(B\). The particle \(P\) can move in the vertical plane containing \(A\) and \(B\). Show that if \(2 b \cosh \varphi=P A+P B\) and \(2 b \cos \theta=P A-P B\), then the kinetic energy of \(P\) is $$ T=\frac{1}{2} m b^{2}\left(\cosh ^{2} \varphi-\cos ^{2} \theta\right)\left(\dot{\theta}^{2}+\dot{\varphi}^{2}\right) $$ Hence find the Lagrangian of the system in terms of \(\theta\) and \(\varphi\).

A particle of unit mass is subject to an inverse-square-law central force $$ \boldsymbol{F}=-\frac{\boldsymbol{r}}{r^{3}} $$ where \(r=|\boldsymbol{r}|\) and \(\boldsymbol{r}\) is the position vector from the origin of an inertial frame. Show that the motion is governed by the Lagrangian $$ L=\frac{1}{2} \dot{r} \cdot \dot{r}+\frac{1}{r} $$ Write down the equations of motion in spherical polar coordinates and show that there are solutions with \(\theta=\pi / 2\) throughout the motion.

A system has Lagrangian \(L=\frac{1}{2} T_{a b} v_{a} v_{b}\) where the \(T_{a b} \mathrm{~s}\) are functions of the \(q_{a} \mathrm{~s}\) alone. Show that $$ \frac{\mathrm{d} L}{\mathrm{~d} t}=0 $$ Show that if \(f\) is any function of one variable, then \(L^{\prime}=f(L)\) generates the same dynamics.

A particle of mass \(m\) is subject to a force \(\boldsymbol{F}\). Obtain the equations of motion in cylindrical polar coordinates.

Two particles, each of mass \(m\), are moving under their mutual gravitational attraction, which is given by the potential \(U=-\gamma m / 2 r\), where \(2 r\) is their separation and \(\gamma\) is a constant. Find the equations of motion in terms of the coordinates \(X, Y, Z, r, \theta, \varphi\), where \(X, Y\), and \(Z\) are the Cartesian coordinates of the centre of mass and \(r, \theta\), and \(\varphi\) are the polar coordinates of one particle relative to the centre of mass.

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